Answer
$$\frac{-1}{2 (1+\sin 2x)}+c$$
Work Step by Step
Given $$ \int \frac{\cos 2 x}{(1+\sin 2 x)^{2}} d x $$
Let
$$u=1+\sin 2 x \ \ \ \ \Rightarrow \ \ \ du =2\cos 2x dx $$
\begin{aligned} \int \frac{\cos 2 x}{(1+\sin 2 x)^{2}} d x&=\frac{1}{2} \int \frac{1}{u^{2}} d u \\
&= \frac{-1}{2u}+c\\
&= \frac{-1}{2 (1+\sin 2x)}+c
\end{aligned}
